7.03 Genetics: Solutions for Problem Set #3 (2000)

Solutions for Problem Set #3 Probleml .-- a. Mutant 1 must be a missense mutation because a full-length protein is produced. Mutant 2 could be either a nonsense mutation or a frameshift mutation because a truncated protein is produced. b. The size difference between mutants 2 and 3 is 5 kDa. The size difference between mutants 3 and 4 is 30 kDa. This means that, in the DNA, the distance between mutants 3 and 4 is about 6-fold greater than the distance between mutants 2 and 3. Thus, the frequency of wild type lysozyme in a cross of 3 with 4 will be 6-fold greater than in a cross of 2 and3. The answer is 12/1000 = 0.012. c. There are two possibilities here. First, you must recognize that the reading frame caused by a +1 frameshift will be different than the reading frame caused by a +2 frameshift. Case #1. There is a stop codon between the two frameshift mutations in the +2 reading frame but not in the + 1 reading frame. Case #2. The amino acid changes between the +1 and -1 frameshifts can be tolerated to yield a functional protein. The amino acid changes between the +2 and -2 frameshifts lead to production of a full length, but non-functional protein product. d. Since both mutations are frameshifts, we know that mutant #2 must occur before codon specifying the amino acid present at the 15 kDa position of the protein. Likewise, mutant #3 must occur before the 20 kDa position. We also know that there is a stop codon at the t5 kDa position in the +1 reading frame and there is a stop codon present at the 20 kDa position in the -1 reading frame. In order for the 2, 3 double mutant to make a full length protein it must be the case that mutant 3 also occurs before the 15 kDa position and restores the proper reading frame ( otherwise i the protein would terminate at the stop codon present in the + 1 reading frame at 15 kDa). So this problem really boils down to calculating the probability that mutant 3 could occur before the 15 kDa position and still produce a 20 kDa product. In other words, what is the probability that over a 5 kDa length of protein no stop codon wiI1 occurat randomin the DNA codingfor that 5 kDa region? To answer this question, we must first calculate how many codons correspond to 5 kDa of protein. Given that each amino acid is 0.110 kDa 5 # of codons- 0.110 - 45.5 Therefore, the probability that no stop codon will occur in 46 random codons is p(rzostop)- 61 46 U'-I" - o.11 Problem 2 a. 5'-CCA-3' b. 5'-CCA-3' - 3'-GGT-5' c. Deamination of cytosine to uracil will result in a net change of C-->T in the DNA and C-->Uin the RNA. Since there are two C's in the tRNA _ anticodon, we must consider each case: 5'-UCA-3' This will suppress a UGA nonsense codon. The DNA sequence is: 5'-TCA-3' 3'-AGT-5' 5'-CUA-3' This will suppress a UAG nonsense codon. The DNA sequence is: 5' -CTA-3' 3'-GAT-5' _ Problem 3 2 a. Some members of the Tn5 insertion collection have transposons that are closely linked to the wild type HIS1 gene. Occasionally one of the phage will accidentally package a portion of the bacterial chromosome that contains both the Tn5 insertion and the wild type HISI gene. When this phage then infects a his]-mutant, His + bacteria result from the foliowing recombination event: Transductant DNA HISI + E.coli Chromosome Tn5 X X . his1- b. The Kan r His + transductants are the result of the cotransduction of HIS1 with Tn5. Therefore, the cotransduction frequency of Tn5 and HISJ is: cotransduction frequency = # His* transductants Total # transductants f" K = 80 100 = 0.8 c. The absence of His +transductants means that HIS2 + and HIS3 + were not cotransduced with Tn5 and HIS1. Therefore, HIS2 and HIS3 must be more than 105 bp away from Tn5 and HISJ. d. There are several possible ways to approach this problem. generating a genomic library and isolating a complementing You could clone his2 by clone. You could then test this clone for the ability to complement, a his3 mutation and sequence the clone to determine which genes are present. An alternative method would be to isolate a strain carrying a Tn linked to the his2 gene. You would then grow P1 on this strain and infect each single mutant. If his2 and his3 are allelic then the cotransduction frequencies should be identical (or very similar). e. All we can really say is that HIS1 and HIS4 are within 105 bp of Tn5. If both genes are on the same side of Tn5 then they are very close together cotransduction frequency ofHIS4 and Tn5 is almost exactly the same as that for HISJ and Tn5. tt is also possible that Tn5 is located between approximately the same distance because the HISI and HIS4 and is from each gene. f. Cross Order A Order B Order C #1" Donor: hisl HIS4 + Tn5 Recipient: HISI + his4 1 4+ Tn5 4+ X X. 1+ 4 His+with 2 crossovers 1 Tn5 1 X X X X. 4 1+ His+ with 4 crossovers Tn5 4+ X X. 1+ 4 His÷ with 2 crossovers #2: Donor: HIS1 + his4-Tn5 Recipient: his1- HIS4 + l 1+ 4- Tn5 . 4- X 1+ Tn5 1+ Tn5 X 4- X X X. 1- 4+ His+ with 4 cross- X X. 4+ 1 His+with 2 cross- X 1 4+ His+ with 2 cross-• overs overs overs We are given in the problem that the number of Kan r His + transductants are greater from cross #1 than cross #2. Therefore, His + must be the result of 2 cross-overs in cross#1 and 4 cross-overs Therefore, in cross#2. the correct order is Order A: HIS1 .... HIS4 ............ Tn5 g. As in part d, you could clone his 1 using a genomic library. determine whether the clone also complements different genes. If the clone complements his4. You could then If not, the mutations are in both his 1 and his4 you would have to sequence the insert to determine how many genes are present on the plasmid. If only a single gene is present then they are allelic. If multiple genes are present, you could subclone each gene individually and test for the ability to complement each mutant.